Formula for counting lattice points in the ordinate set of a function

For a nonnegative function $f$ defined on an interval $[a,b]$ define the set

$S = \{ (x,y) \mid a \leq x \leq b, \ 0 < y \leq f(x) \}.$

(i.e., the region enclosed by the graph of the function and the $x$ -axis between the vertical lines at $x=a$ and $x=b$ ). Then prove

$\text{\# of lattice points in } S = \sum_{n=a}^b [f(n)],$

where $[f(n)]$ is the greatest integer less than or equal to $f(n)$ .

We can help ourselves by drawing a picture to get a good idea of what is going on, then turn that intuition into something more rigorous. The picture is as follows:

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In the picture, we can see the number of lattice points in the ordinate set of $f(x)$ (not including the $x$ -axis since the question stipulates $S$ contains the points $0 < y \leq f(x)$ ). At each integer between $a$ and $b$ , we count the number of lattice points beneath $[f(x)]$ , the greatest integer less than or equal to $f(x)$ . Then we need to turn this intuition from the picture into a proof:

Proof. Let $n \in \mathbb{Z}$ with $a \leq n < b$ . We know such an $n$ exists since $a,b \in \mathbb{Z}$ with $a < b$ . Then, the number of lattice points in $S$ with first element $n$ is the number of integers $y$ such that $0 < y \leq f(n)$ . But, by definition, this is $[f(n)]$ (the greatest integer less than or equal to $f(n)$ ). Summing over all integers $n, \ a \leq n \leq b$ we have,